Relativity1905 - Albert Einstein:

• Brownian motion

fi atoms.

• Photoelectric effect.

fi Quantum Theory

• “On the Electrodynamics of Moving Bodies”

fi The Special Theory of Relativity

The Luminiferous Ether

Hypothesis: EM waves (light) travelthrough some medium - The Ether

Speed of light: c = 3 x 108 m/s w.r.t fixed ether.

The earth moves at v = 3 x 104 m/sw.r.t fixed ether.

fi Speed of light w.r.t earth should depend on direction.

The Michelson-MorleyExperiment

An interferometer

*LightSource

Mirror A

Mirror B

Telescope

Beam A

Beam B

1/2 Silvered Mirror

The interference fringes should shift.

But no effect was observed!

What was wrong?

The Lorentz-Fitzgerald Contraction

Suppose that the ether squashes anyobject moving through it?

To counteract the change in light speed,we need: d’ = d √1 - v2/c2

Galilean Transformations.

Ky

x

z

K’y’

x’

z’

t’ = tz’ = zy’ = yx’ = x - vt

In frame K, two charges at rest. Force isgiven by Coulomb’s law.

Ky

x

z

Q1

Q2

K’y’

x’

z’

In moving frame K’, two charges are moving.Since moving charges are currents,Force is Coulomb + Magnetism.

Principle of relativity:

“The laws of nature are the same in allinertial reference frames”

Something is wrong!

• Maxwell’s Equations?

• The Principle of Relativity?

• Gallilean Transformations?

Einstein decidedfi Galilean Transformations are the

problem.

Einstein’s two postulates:

1. The principle of relativity is correct.The laws of physics are the same in allinertial reference frames.

2. The speed of light in vacuum is thesame in all inertial reference frames(c = 3 x 108 m/s regardless of motion ofthe source or observer).

The second postulate seems to violateeveryday common sense!

Rocket Light pulse Observerv=0.5 c v=c

Einstein says: observer measures the lightas traveling at speed c, not 1.5c.

Gedanken ExperimentsA light clock:

It ticks every Dt = 2 w/c seconds.One can synchronize ordinary clocks withit.

w

mirror

mirror photocell

Time Dilation

OG: Observer on Ground

w

OT: Observer on Truck

OT’s clock as seen from the ground:

c = 3 x 108 m/s

(ct/2)2 - (vt/2)2 = w2

w

v

w

vt/2

ct/2

Time for one round trip of light, as seenfrom the ground:

t = (2 w/c) √1 - v2/c2

For v = 0.6c, t = (2 w/c) x 1.25

All of OT’s processes slow down comparedto OG as seen by OG.

Similarly,

All of OG’s processes slow down comparedto OT as seen by OT.

Length Contraction

OG: Observer on Ground

w

OT: Observer on Truck

Device on truck makes mark on trackeach time clock ticks.

As seen from ground:

Distance between marks= (time between ticks) x v

= [(2 w/c) √1 - v2/c2 ] v

w

v

As seen from truck:

Distance between marks= (time between ticks) x v

= (2 w/c) v

(To the person on the truck the timebetween ticks is (2 w/c).)

(Distance measured on truck)

= √ 1 - v2/c2

x (distance measured on ground)

As seen from a moving frame, restdistances contract.

(L-F contraction)

SimultaneityEvents occur at a well defined position anda time (x,y,z,t).

But events that are simultaneous (same t)in one inertial frame are not necessarilysimultaneous in another frame.

The light from the two flashes reach OGat the same time. He sees them assimultaneous.

d d

A B

OT passes OG just as the lights flash.

v

But light from B reaches OT first. Sinceboth light beams started the samedistance from her, and both travel atspeed c, she concludes that B must haveflashed before A.

v

cc

Lorentz Transformations

• Flashbulb at origin just as both axes are coincident.• Wavefronts in both systems must be spherical:

x2 + y2 + z2 = c2t2 and

x’2 + y’2 + z’2 = c2t’2

• Inconsistent with a Galilean transformation• Also cannot assume t=t’.

y

x

z

y’

x’

z’

v

Assuming:

• Principle of relativity

• linear transformation (x,y,z,t) -> (x’,y’,z’,t’)

Lorentz Transformations (section 2.4)

x’ = g ( x - v t )

y’ = y

z’ = z

t’ = g ( t - v x / c2 )

With g = 1 / √1 - v2/c2 .

(Often also define b = v / c . )

Time Dilation (again)Proper time: time T0 measured betweentwo events at the same position in aninertial frame.

OG’s clock: T0 = t2 - t1, (x2-x1=0)

OT’s clock: T’ = t’2 - t’1

t’2 - t’1 = g (t2 - t1 - v/c2 (x2-x1) )

T’ = g T0 > T0

Clocks, as seen by observers moving at arelative velocity, run slow.

v

OG

OTOT’s friend

Length Contraction (again)Proper length: distance L0 between pointsthat are at rest in an inertial frame.

OT on truck measures its length to be L0 = x’2 - x’1. This is its proper length.OG on ground measures its length to beL = x2 - x1, using a meter stick at rest(t2 = t1).

ThenL0 = x’2 - x’1 = g (x2 - x1 - v (t2 - t1))

= g L

OG measures L = L0/ g < L0.

Truck appears contracted to OG.

vx’1 x’2

x1 x2

An applicationMuon decays with the formula:

N = N0 e-t/t

N0 = number of muons at time t=0.N = number of muons at time t secondslater.t = 2.19 x 10-6 seconds is mean lifetime ofmuon.

Suppose 1000 muons start at top ofmountain d=2000 m high and travel atspeed v=0.98c towards the ground. Whatis the expected number that reach earth?

Time to reach earth:t = d/v = 2000m/(0.98 x 3 x 108 m/s) = 6.8 x 10-6 s

Expect N = 1000 e-6.8/2.19 = 45 muons.

But experimentally we see 540 muons!What did we do wrong?

Time dilation: The moving muon’s internalclock runs slow. It has only gone through

t’ = 6.8 x 10-6 √1 - 0.982 s = 1.35 x 10-6 s

So N = 1000 e-1.35/2.19 = 540 muons survive.

Alternate explanation: From muon’sviewpoint, the mountain is contracted.Get same result.

Addition of velocitiesGalilean formula (u=u’+v) is wrong.

Consider object, velocity u’ as seen inframe of OT who is on a truck moving withvelocity v w.r.t the ground.

What is velocity u of the object asmeasured by OG on the ground?

Recall u = Dx/Dt, u’ = Dx’/Dt’.Inverse Lorentz transformation formulae:

Dx = g ( Dx’ + v Dt’ )

Dt = g (Dt’ + v Dx’ / c2 )

v

u’u?

Dx g ( Dx’ + v Dt’ ) u = =

Dt g (Dt’ + v Dx’ / c2 )

u’ + v u = 1 + v u’/c2

For u’ and v much less than c:

u ≈ u’ + v

Velocities in y and z directions are alsomodified (due to t’≠t, see section 2.6)

Examples:

Rocket Light pulse Observerv=0.5 c u’=c

Observer sees light move at

0.5c + c u = = c 1+(0.5c)(c)/c2

Light moves at c=3x108 m/s in all frames.

Rocket Projectilev=0.8 c u’=0.5c Observer

Observer sees projectile move at

0.5c + 0.8c u = = 0.93c 1+(0.5)(0.8)

Massive objects always move at speeds < c.

The Twin ParadoxSuppose there are two twins, Henry andAlbert. Henry takes a rocket ship, goingnear the speed of light, to a nearby star,and then returns. Albert stays at home onearth.

Albert says that Henry’s clocks arerunning slow, so that when Henry returnshe will still be young, whereas Albert is anold man.

But Henry could just as well say thatAlbert is the one moving rapidly, so Albertshould be younger after Henry returns!

Who is right?

The first scenario is the correct one.

The situation is not symmetric, becausethe rocket has to decelerate, turn aroundand accelerate again to return to earth.Thus, Henry is not in an inertial framethroughout the trip. He does returnyounger than Albert.

Relativistic Doppler EffectLight source and observer approach eachother with relative velocity, v.Light is emitted at frequency n0.

Observer sees light at a higher frequency:

√1 + b n = n0 with b = v/c √1 - b

• If source is receding, the formula still holds but now b is negative.

We know that the universe is expanding,because light from distance galaxies isred-shifted, indicating motion away fromus.

v

n0

n

Relativistic MomentumRequirement: momentum is conserved in allinertial frames.Assume: p = m v.

Elastic scattering in c-o-m frame:

px: mu + m(-u) = 0

Transform to frame of A:

-2upx: 0 + m( ) ≠ m(-2u) 1+u2/c2

It doesn’t work!

AB

before afterA

B

AB

A

B

Relativistic momentum:

mv p = g m v = √1-v2/c2

Relativistic Kinetic Energy:

K = (g - 1) mc2

1 = ( - 1) mc2

√ 1-v2/c2

For small velocities, v/c << 1:

K = ( 1 + 1/2 (v/c)2 + ... - 1) mc2

≈ 1/2 m v2

For large velocities v c:

K •

Massive objects always travel at speedsless than c.

Relativistic EnergyAccording to Einstein, even a mass at resthas energy:

E0 = m c2 (rest energy)

Thus, the total energy of a moving objectis E = K + E0

= (g-1) mc2 + mc2

= g mc2

It is straightforward to show:

E2 - p2c2 = m2c4

For a massless particle (e.g. a photon):

E = |p| c

In general

|p| c2 g mv c2 v = = E g mc2

For a massless particle this gives

v = c

Massless particles travel at the speed oflight c.